Question

(V) DATA AND RESULTS: 3 Pressure. P (kPa) Volume, V (L) PxV_(kPa*L) 1P. (1/kPa) 114.821 1.5 172.2315 0.0087092082 117.757 0.0084920642 141.3084 108.5544 120.616 09 0.008290774 123.346 0.6 74.0076 0.0081072755 126.048 04 50.4192 0.0079334857

Answer 1

(1) Boyle's Law: It states that absolute pressure and volume of a given mass of ideal gas are inversely proportional, provided the temperature remains unchanged and the gas molecules donot excake out of container s 2 / Pa cost a t constant Tempera - ture Conditions ② T in Kelvin must be constant The amount of gas should not change Systern should be isolated from environment. ② (1) Since from Boyle's Law ev= const. ve costi & Graph between V vs Ye Should be a slope of straight line of constant Slope. Since from graph we can see that, all points passes through a single straight line, Therefore Boyle's law Confirmed.

1.8 1.6 - 1.4 - Volume (Lit) 0.4 - 0.2 + 7.8 7.9 8.0 8.6 8.7 8.8 8.1 8.2 8.3 8.4 8.5 1/Pressure (inv Pa)

(11) from Univeral Gran canation ev=nRT at constant tomberature above law becomes Boyle's law PV = nRT = const » (Boyle's low) - Pivi = P₂V, Given P,= 760 mm Hg V,= 10m2 P₂ = ? V₂ = 5m2 = P = Pivi - 760010 pressure after = 1520 mmitta compression is f = 1520 mmity Given V,= 22.4 Lit P = 1 atmospheric pressure P2 = 2 atmospheric Pressure using Boyle's law Similar to Problem (111) Pr= const =) P. V = P2 V /2 = Piv V2 = 22 22.4 x 10th 2aton V - 11.2 Lit volume will be New = 11:22it.

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